Question

A capillary tube of radius $r$ is immersed vertically in a liquid such that liquid rises in it to height $h$ (less than the length of the tube). Mass of liquid in the capillary tube is $m$. If radius of the capillary tube is increased by $50 \%$, then mass of liquid that will rise in the tube, is

• A. $\frac {2}{3} m$
• B. m
• C. $\frac {3}{2} m$
• D. $\frac {9}{4} m$

Answer 1

Answer: (C)

$\frac {3}{2} m$

Answer 2

$h=\frac{2 T \cos \theta}{r \rho g}$ $\Rightarrow h \propto \frac{1}{r}$ $\Rightarrow \frac{h_{2}}{h_{1}}=\frac{r_{1}}{h_{1}}=\frac{2}{3}$ $\left(\therefore r_{1}=r_{2}=r+50 \%\right.$ of $\left.r=\frac{3}{2} r\right)$ New mas $m_{2}=\pi r_{2}^{2} h_{2} \rho$ $=\pi\left(\frac{3}{2} r_{1}\right)^{2}\left(\frac{2}{3} h_{1}\right) \rho \frac{3}{2} m$